Chapter 5.4, Problem 15E

To determine

To find: use part 1 of the fundamental theorem of calculus to find the derivative of the function.

Answer to Problem 15E

The derivative of the function is $\sqrt{\text{tanx+}\sqrt{\text{tanx}}}·{\text{sec}}^2\text{x}$ .

Explanation of Solution

Given information: The function is $\sqrt{\text{tanx+}\sqrt{\text{tanx}}}·{\text{sec}}^2\text{x}$

Let’s remind of the fundamental theorem of calculus part 1:

The fundamental theorem of calculus part 1: If f is continuous on $[\text{a,b}]$ then the function of g defined by

  $\text{g(x)=}{\displaystyle {\int }_{\text{a}}^{\text{x}}\text{f(t)}}\text{ dt}$ where $\text{a}\le \text{x}\le \text{b}$ is continuous on $[\text{a,b}]$ and differentiable on $(\text{a,b})$ and $\text{<msup><mo>g</mo><mo>′</mo></msup><mo>(x)=f(x)</mo>}$ .

We have, $\text{y=}{\displaystyle {\int }_0^{\text{tan x}}\sqrt{\text{t+}\sqrt{\text{t}}}}\text{ dt}$

Since the upper limit of integration is not y , we apply the chain rule

Let , $\text{y=f(u) , u=tanx}$

Then, $\text{u=tanx}$ , ${\text{<msup><mo>u</mo><mo>′</mo></msup><mo>=sec</mo>}}^2\text{x}$

Consider the new function

  $\text{f(u)=}{\displaystyle {\int }_0^{\text{u}}\sqrt{\text{t+}\sqrt{\text{t}}}}\text{ dt}$

Differentiation with respect to u,

  $\Rightarrow \frac{\text{d}}{\text{du}}\text{f(u)=}\frac{\text{d}}{\text{du}}[{\displaystyle {\int }_0^{\text{u}}\sqrt{\text{t+}\sqrt{\text{t}}}}\text{ dt]}$

  $\Rightarrow \frac{\text{d}}{\text{du}}\text{f(u)}$

  $=\sqrt{\text{u+}\sqrt{\text{u}}}$

  $\Rightarrow \text{<msup><mo>f</mo><mo>′</mo></msup><mo>(u)</mo>}=\sqrt{\text{u+}\sqrt{\text{u}}}$

Now $\text{y=f(u)}$

  $\Rightarrow \text{<msup><mo>y</mo><mo>′</mo></msup><mo>=[f(tanx)u<msup><mo>]</mo><mo>′</mo></msup></mo>}$

  $\Rightarrow \text{<msup><mo>y</mo><mo>′</mo></msup>=<msup><mo>f</mo><mo>′</mo></msup><mo>(tanx)</mo>}·\text{<msup><mo>u</mo><mo>′</mo></msup>}$

  $\text{<msup><mo>y</mo><mo>′</mo></msup>=}\sqrt{\text{tanx+}\sqrt{\text{tanx}}}·{\text{sec}}^2\text{x}$

Hence, the derivative is $\sqrt{\text{tanx+}\sqrt{\text{tanx}}}·{\text{sec}}^2\text{x}$ .